A prism of an equilateral triangle base , its height is triple the length of its base if the length of its base increases by rate 0.01 cm/sec . Find the rate of increase of volume of the prism at length of base 5 root3 cm

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- Mar 3rd 2013, 07:44 AMmido22rate of change according to time problem
A prism of an equilateral triangle base , its height is triple the length of its base if the length of its base increases by rate 0.01 cm/sec . Find the rate of increase of volume of the prism at length of base 5 root3 cm

- Mar 3rd 2013, 10:36 AMHallsofIvyRe: rate of change according to time problem
I'm not sure what you mean by "the length of its base". The obvious thing is the length of a

**side**of the triangle but another possibility is the length of an altitude.

First, can you write the formula for volume of such a prism with base side length s and height h? - Mar 3rd 2013, 12:06 PMmido22Re: rate of change according to time problem
the base length means -->the length of a side of the triangle

here is my attempt :

volume of prism = area of base * height .....assume that base length = x ,, height of prism = h = 3x

so

$\displaystyle v= (\frac{1}{2} \times X \times \frac{\sqrt{3}}{2}X )\times 3X$

$\displaystyle v = \frac{3 \sqrt{3}}{4} \times x^3 $by taking derivative with respect to time

so

$\displaystyle \;\frac {dv}{dt} = \frac{9 \sqrt{3}}{4} \times x^2 \times \frac{dx}{dt}$ by substituting i found that

$\displaystyle \frac {dv}{dt} = \frac{27 \sqrt{3}}{16} $

is that right?????(Worried)